Question: Two cars are driving away from an intersection in perpendicular directions. The first car's velocity is $7$ meters per second and the second car's velocity is $3$ meters per second. At a certain instant, the first car is $5$ meters from the intersection and the second car is $12$ meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{71}{13}$ (Choice B) B $\dfrac{99}{13}$ (Choice C) C $13$ (Choice D) D $\sqrt{58}$
Solution: Setting up the math Let... $a(t)$ denote the distance between the first car and the intersection at time $t$, $b(t)$ denote the distance between the second car and the intersection at time $t$, and $c(t)$ denote the distance between the two cars at time $t$. $a(t)$ $b(t)$ $c(t)$ We are given that $a'(t)=7$ and $b'(t)=3$. We are also given that $a(t_0)=5$ and $b(t_0)=12$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $[a(t)]^2+[b(t)]^2=[c(t)]^2$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{1}{c(t)}\left[a(t)a'(t)+b(t)b'(t)\right]$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=5$ and $b(t_0)=12$, we can find that $c(t_0)=13$. Let's plug ${c(t_0)}={13}$, ${a(t_0)}={5}$, ${a'(t_0)}={7}$, $C{b(t_0)}=C{12}$, and ${b'(t_0)}={3}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{1}{{c(t_0)}}\left[{a(t_0)}{a'(t_0)}+C{b(t_0)}{b'(t_0)}\right] \\\\ &=\dfrac{1}{({13})}\left[({5})({7})+(C{12})({3})\right] \\\\ &=\dfrac{71}{13} \end{aligned}$ In conclusion, the rate of change of the distance between the two cars at that instant is $\dfrac{71}{13}$ meters per second. Since the rate of change is positive, we know that the distance is increasing.